How does size of the dataset depend on VC dimension of the hypothesis class?
Given a hypothesis set $H$, the set of all possible mappings from $X\to Y$ where $X$ is our input space and $Y$ are our binary mappings: $\{-1,1\}$, the growth function, $\Pi_H(m)$, is defined as the maximum number of dichotomies generated by $H$ on $m$ points. Here a dichotomy is the set of $m$ points in $X$ that represent a hypothesis. A hypothesis is just a way we classify our points. Therefore with two labels we know,

$$\Pi_H(m)\leq 2^m$$

This is just counts every possible hypothesis. The VC dimension is then the largest $m$ where $\Pi_H(m)=2^m$.

Consider a 2D perceptron, meaning our $X$ is $\mathbb{R}^2$ and our classifying hyperlane is one-dimensional: a line. The VC dimension will be 3. This is because we can shatter (correctly classify) all dichotomies for $m=3$. We can either have all points be the same colour, or one point be a different colour - which is $2^3=8$ dichotomies. You may ask what if the points we are trying to classify are collinear. This does not matter because we are concerned with resolving the dichotomies themselves, not the location of the points. We just need a set of points (wherever they may be located) that exhibits that dichotomy. In other words, we can pick the points such that they maximize the number of dichotomies we can shatter with one classifying hyperplane (a triangle): the VC dimension is a statement of the capacity of our model.

To make this clear, consider $m=4$. We can represent the truth table of the XOR gate as a dichotomy but this is not resolvable by the perceptron, no matter where we choose the location of the points (not linearly separable). Therefore, we can resolve a maximum of 8 dichotomies, so our VC dimension is 3. In general, the VC dimension of perceptrons is $d+1$ where $d$ is the dimension of $X$ and $d-1$ is the dimension of the classifying hyperplane.